Chapter 7.2, Problem 37E

To determine

To find: that the sample size would be required to reduce the standard deviation of the sampling distribution to one-half the value.

Answer to Problem 37E

120

Explanation of Solution

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

It requires reducing the standard deviation to one half of the value.

  $\frac{{\sigma }_{\widehat{p}}}{2}=\frac{1}{2}\sqrt{\frac{p\left(1-p\right)}{n}}=\sqrt{\frac{1}{2}\cdot \frac{p\left(1-p\right)}{n}}=\sqrt{\frac{p\left(1-p\right)}{4n}}$

It observed that $n$ were replaced by $4n$ in the equation of the standard deviation, when the standard deviation is reduced to one-half of its value, which implies that sample size need to be multiplied by 4.

The sample size in the previous exercise was 30.

  $4n=4\left(30\right)=120$

It implies that if using his sample size of 120, then the standard deviation of the sampling distribution of the sample proportions is reduced to one-half of its original value.